$\dfrac{dy}{dx}=x+4y-2$ Is $y=-\dfrac{x}{4}+\dfrac{3}{4}$ a solution to the above equation? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Explanation: In order to find whether $y=-\dfrac{x}{4}+\dfrac{3}{4}$ is a solution, we need to substitute it into the equation and see if we get equivalent expressions on each side of the equal sign. In addition to substituting for $y$, we need to find the corresponding $\dfrac{dy}{dx}$ expression to substitute into the equation: $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{d}{dx}\left[-\dfrac{x}{4}+\dfrac{3}{4}\right] \\\\ &=-\dfrac{1}{4} \end{aligned}$ Now we substitute ${y=-\dfrac{x}{4}+\dfrac{3}{4}}$ and ${\dfrac{dy}{dx}=-\dfrac{1}{4}}$ into the equation: $\begin{aligned} {\dfrac{dy}{dx}}&=x+4{y}-2 \\\\ {-\dfrac{1}{4}}&\stackrel{?}{=}x+4\left({-\dfrac{x}{4}+\dfrac{3}{4}}\right)-2 \\\\ -\dfrac{1}{4}&\stackrel{?}{=}x-x+3-2 \\\\ -\dfrac{1}{4}&\neq 1 \end{aligned}$ We did not obtain equivalent expressions on each side. In conclusion, no, $y=-\dfrac{x}{4}+\dfrac{3}{4}$ is not a solution to the differential equation.